A block of mass m is placed on a surface having a vertical cross-section given by . If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is

Question

A block of mass m is placed on a surface having a vertical cross-section given by $y = x^{2} / 4$ . If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is [2024].

Smart Achievers · Accepted Answer

(1)

$f_{s} = m g \sin θ$ ...(1)

$N = m g \cos θ$ ...(2)

Now, $f_{s} \leq μ N$

$m g \sin θ \leq μ m g \cos θ$

$\tan θ \leq μ$ ...(3)

According to question, $y = \frac{x^{2}}{4}$

$\tan θ = \frac{d y}{d x} = \frac{1}{4} \cdot 2 x = \frac{x}{2}$

Put in (3) $\frac{x}{2} \leq μ \Rightarrow x \leq 2 (0.5)$

$\Rightarrow x \leq 1$ $So y \leq \frac{1^{2}}{4} \Rightarrow y_{\max} = \frac{1}{4} = 0.25$

Solution

Q.
A block of mass m is placed on a surface having a vertical cross-section given by $y = x^{2} / 4$ . If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is [2024].

1 1/4 m

2 1/2 m

3 1/6 m

4 1/3 m

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